Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19907 Accepted Submission(s): 4472
There is exactly one node, called the root, to which no directed edges point.
Every node except the root has exactly one edge pointing to it.
There is a unique sequence of directed edges from the root to each node.
For example, consider the illustrations below, in which nodes are represented by circles and edges are represented by lines with arrowheads. The first two of these are trees, but the last is not.
In this problem you will be given several descriptions of collections of nodes connected by directed edges. For each of these you are to determine if the collection satisfies the definition of a tree or not.
package 并查集; import java.util.Scanner; public class hdu_1325 { static int [] father; static int [] indegree; static boolean [] flag; public static void main(String[] args) { Scanner sc = new Scanner(System.in); int k = 1; while(true){ boolean istree=true; father = new int [1000]; indegree = new int [1000]; flag = new boolean [1000]; int x = sc.nextInt(); int y =sc.nextInt(); if(x<0&&y<0) break; if(x==0&&y==0){ //空树也是树 System.out.println("Case "+(k++)+" is a tree."); continue; } for(int i=1;i<1000;i++){ father[i] = i; } while(true){ if(x==0&&y==0) break; int a = find(x); int b = find(y); flag[x] = true; flag[y] = true; if(a!=b) { father[a] = b; indegree[y]++; //入度加一 } if(a==b) istree = false; //成环 x = sc.nextInt(); y = sc.nextInt(); } int ans = 0; for(int i=1;i<1000;i++){ if(indegree[i]>1) { istree = false; //度大于一 } if(flag[i]&&father[i]==i) ans++; //这个点被标记过了并且自己还是一棵树的根 if(ans>1) istree = false; //森林 } if(istree) System.out.println("Case "+(k++)+" is a tree."); else System.out.println("Case "+(k++)+" is not a tree."); } } private static int find(int x) { if(x==father[x]) return x; return find(father[x]); } }